GATE Mathematics Mock Test - 5 - GATE Mathematics MCQ

With 'dare' as a main verb, to-infinitive will be used. Option (2) is the correct answer.

Problem figure (5) is the same as figure (1). So, the answer figure will be the same as figure (2)

Interest paid by Kishan Singh = Rs. 48,000

Now,

48 = 6x + 40 - 4x

2x = 8

x = 4 years

It means Kishan Singh availed the discount after 4 years.

option (3) is correct.

Let number of students who play hockey = y

Let number of students who play both cricket and hockey = C

Now as per information in questions:

x = 2 (y- C) …………(i)

y = 2 (x- C) ………….(ii)

Putting value of y from equation (ii) in equation (i)

x= 2 [2 (x- C)- C] = 2 [2x - 3C]

= 4x- 6C

or ax = 2C (iii)

Now, putting value in equation (ii)

y= 2c ... (iv)

Sum total of those students who play either cricket or hockey or both;

S, = (x - C)+ (y- C)+ C = (2C- C)+ (2C- C)+ C

= C + C = 3C

Some total of those who play only cricket and only hockey

S2 = - + (y- C)= (2C- C)+ (2C- C)= C + C = 2C

Let 

Therefore , option(a) is incorrect.

For option (b) , let 

But 

Therefore option (b) is correct

For optin (c0 and (d)

Also in T.

Therefore option (c) and (d) are incorrect.

y" -2y' + y = e^-x

Where a₁ = -2, a₂ = 1 are constants and b (x) = e^-x is a continuous function an [0,∞)

Now, y(x) = e^x + xe^x +1/4e^-x is one of the solutions of given ODE

Since b(x) =e^-x is bounded on [0,∞)

but y(x) is unbounded on [0,∞) because 

Both statement are false

Hence, series  converges and sum is 5e.

Closure of  Which is uncountable

Becausw x√2 is an irrational number for 

Number of elements of order d in Z_n where ���� is (d).

Therefore, number of elements of order p in Z_p²_q = (p) = p-1

We can write general team for r ≥ 2

f : ℝ→ℝ is an odd function, so

f(-x) = -f(x) ∀ x ∈ ℝ

differeniating both side, we have

-f'(x) = -f'(x) i.e. f'(-x) = f'(x)

Given 

⇒ (D³ + 3D² + 3D + 1)y = 0

A.E. is m³ + 3m² + 3m + 1 = 0

⇒ (m + 1 )³ = 0

⇒ m = -1

therefore general solution 

y(x) = (c₁ + c₂x + c₃x²)e^-x

The given equation is,

Differentiating equation (1) with respect to x, we get

Now, again differentiating equation (1) with respect to y, we get

Hence, the correct answer is 0.

The matrix form of the given system of equations is,

The given system of equations will have a unique solution if and only if the coefficient matrix is non-singular.

Performing , we get,

Performing , we get

Therefore the coefficient matrix will be non-singular if and only if,

Thus the given system will have a unique solution if 

Showing that given equations are inconsistent in this case. Thus if  

= -3.5 no solution exists.

Hence, the correct answer is −3.5

0

In the question, y : x to infinity x : 0 to infinity.
Now changing the order of integration:
y = x
y tends to infinity
y : 0 to infinity x : 0 to y

Hence, the correct answer is 1.

Hence, the correct answer is 14.

Since let 

By L hospital rule 

Given

replace x → -x

from(i) and (ii)

therefore,

from (iii) and (iv)

I = 0

Given a_n = 3^-n a_n-1

Let 

 convergent if |z| < √3 and divergent if |z| > √3

So, radius of convergence is √3.

(I) 

Here f (x.y)  = 

hence (1. 1) is not in the Domain of function, hence along the curvey = x limit do not exist. So limit (I) do not exist.

(II)  

= 

For option (a)

We know that if the system Ax = b of m linear equation in n unknowns is consistent then the solution set Ax = b has n-r+1 linearly independent solution, where r is the rank of matrix A.

Thus the dimension of the solution set of Ax = b is n-r+1.

⇒ option (a) is not ture.

For option (b)

Let 

The system Ax = b is consistent.

We know that if the system  is consistent and rank (A) < n, then system has infinite solutions.

⇒ option (b) is not true

For Option (c)

Clearly Rank (A) ≠ Rank ([A : b])

⇒ This system has no solution.

⇒ Option (c) is not true.

Given

x⁵ + x³ -2 = 0

Let f(x) = x⁵ + x³ -2 

⇒ f(x) = 5x⁴ +3x² ≥0 ∀x ∈ ℝ ⇒ f is increasing and degree of f is odd

Therefore, f has only one real root.

Hence no. of real root is 1.

Re-wrirte the equation as

Compare with general form y" + py' +Qy = 0, we have p(t) = 

Now using W (1) = 1, we have 1 =ce i.e. c = 1/e

⇒ W(t) = t²e^t-1

Let us consider A_n = 

then A_n is open for each n ∈ ℕ. But

 is not open.

We have

⇒ for every closed path .